# 47261948 Simulation Analysis of Lighting Switching Impulse Voltages

August 13, 2017 | Author: peppeto373137 | Category: High Voltage, Electrical Network, Switch, Electric Power System, Insulator (Electricity)

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imulation Analysis of Lighting Switching Impulse Voltages...

#### Description

Simulation Analysis of Lighting Switching Impulse Voltages Ahmad Shaiful Amiruddin Taher(1071118187) Irwan Ramli(1081116195) Ariff Musa(1051105968) December 27, 2010 Abstract This report serve the purpose of the simulation and analysis of lighting and switching impulse voltages as part of our High Voltage Engineering subject demand. The objectives is to simulate switching and lightning impulse voltages, to study the behavior of circuit parameters on the impulse wave shape (tf &tt ) and to be able to simulate and compute the required impulse peak voltage for given front and tail times by varying circuit parameters.

Contents 1 Introduction

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2 Question 1 2.1 Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 MATLAB® simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Question 2 3.1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Matlab® simulations . . . . . . . . . . . . . . . . . . 3.3 the Matlab program to compute for lightning impulse 3.4 Analysis and verification of the result . . . . . . . . .

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Figure 1: This is the caption for the picture.

1

Introduction

Power systems equipment must withstand not only the rated voltage (Vm), which corresponds to the highest voltage of a particular system, but also overvoltages. Accordingly, it is necessary to test high voltage equipment during its development stage and prior to commissioning. The magnitude and type of test voltage varies with the rated voltage of a particular apparatus. The standard method of measurement of high voltage and the basic techniques for application to all types of apparatus for alternating voltages, direct voltages, switching impulse voltages and lightning impulse voltages are laid down in the relevant national and international standards • Testing with lightning impulse voltages. Lightning strokes terminating on transmission lines will induce steep rising voltages in the line and set up travelling waves along the line and may damage the systems insulation. The magnitude of these overvoltages may reach several thousand kilovolts, depending upon insulation. Exhaustive measurements and long experience have shown that lightning overvoltages are characterized by short front duration, ranging from a fraction of a microsecond to several tens of microseconds and then slowly decreasing to zero. The standard impulse voltage has been accepted as an aperiodic impulse that reaches its peak value in 1.2s and then decreases slowly (in about 50s) to half its peak value. In addition to testing equipment, impulse voltages are extensively used in research laboratories in the fundamental studies of electrical discharge mechanism, notably when the time to breakdown is of interest. • Testing with switching impulse voltages. Transient overvoltages accompanying sudden changes in the state of power systems, e.g. switching operations of faults, are known as switching impulse voltages. It has become generally recognized that switching impulse voltages are usually the dominant factor affecting the design of insulation in high voltage power system for rated voltages of about 300kV and above. Accordingly, the various international standards recommend that equipment designed for voltages above 300kV be tested for switching impulses. Although the waveshape of switching overvoltages occurring in the system may vary widely, experience has shown that for flashovevr distances in atmospheric air of practical interest the lowest withstand values are obtained with surges with front times between 100 and 300s. Hence, the recommended switching surge voltage has been designated to have a front time of about 250s and half value time of 2500s.

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Question 1

A Cockcroft Walton type voltage tripler circuit has C1 = C2 = 0.025 µF and C3 = 0.15 µF.The supply voltage is 105 sin ωt kV, where ω= 314. If the load current is 25 mA, determine: (a) ripple voltage (b) voltage drop (c) average output voltage (d) ripple factor (e) simulate the circuit using MATLAB simulink to verify above calculations

2.1

Calculations

• ripple voltage !

1 2 I1 + = f C1 C3 #  " 25mA 1 2 = + 50 0.025µF 0.15µF = 43.33kV

∆V



(1)

• Voltage drop ∆V δV + 2

!

I1 1 1 1 = + + f C2 C2 2C3 # !" 1 1 25mA 1 + + = 314 0.025µF 0.025µF 2(0.01µF ) 2π = 41.6878kV 

(2)

• Average output voltage ∆V = 3Vs(max) − δV + 2 = 3(105kV − 41.6878kV ) = 273.3122kV 

Vav

3



(3)

• Ripple factor 

RF =

δV +

∆V 2



3Vs(max) 41.6878kV = 3(105kV ) = 0.1323

2.2

MATLAB® simulations

• circuit diagram

4

(4)

• graph plotted

Figure 2: Graph plotted from Matlab® simulink

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Question 2

Figure 3 shows the equivalent circuit of a single stage impulse generator with input voltage of 300 kV. The value of each component used in this circuit is C1= 7 nF, C2= 300 pF, R1=1.6 kΩ, R2= 6.5 kΩ. (a) Obtain the Vout(t) and its efficiencies. (b) Using the MATLAB® simulink, plot the waveform. (c) Using the file in the Energy Systems Lab (name of the file is IGA), modify the Matlab program to compute for lightning impulse wave form (d) Analyse and verify the results

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Figure 3: Single stage impulse generator with input voltage of 300 kV

3.1

Solutions

• Obtain the Vout(t) and its efficiencies. Vc R1 C2 300kV = (1.6kΩ)(200pF ) = 6.25 × 1011

K =

R1 C2 + R2 C2 + R2 C1 R1 R2 C1 C2 (1.6kΩ)(200pF ) + (6.5kΩ)(300pF ) + (6.5kΩ)(7nF ) = (1.6kΩ × 6.5kΩ × 7nF × 300pF ) = 2.1946 × 106

α =

β =

1 R1 R2 C1 C2

1 (1.6kΩ × 6.5kΩ × 7nF × 300pF ) = 4.5788 × 1010

=

Vout =

K s2 + αs + β

Vout =

6.25 × 1011 [s2 + 2.1946 × 106 s + 4.5788 × 1010 ]

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[s2 + 2.1946 × 106 s + 4.5788 × 1010 ] = 0 1 2.1946 × 106 4.5788√× 1010 −b ± b2 − 4ac S = 2a

B(s) a b c

= = = =

√ b2 − 4ac S = 2a q −2.1946 × 106 ± (2.1946 × 106 )2 − 4(1)(4.5788 × 1010 ) S = 2(1) = −21066, −2173534 −b ±

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Vout = 6.25 × 10



B A + s + 21066 s + 2173534



by comparison with:

Vout =

6.25 × 1011 [s2 + 2.1946 × 106 s + 4.5788 × 1010 ]

we get A+B = 0 2173534A + 21066B = 1 solve for (5) and (6): 1 = 4.6458 × 10−7 2152468 −1 B = = −4.6458 × 10−7 2152468 " # −7 4.6458 × 10−7 11 4.6458 × 10 Vout = 6.25 × 10 − s + 21066 s + 2173534 A =

7

(5) (6)

by taking Laplace Transform. Vout = 290.3625(e−21066t − e−2173534t )kV dVout d = 290.3625(e−21066t − e−2173534t )kV = 0 dt dt

−21066e−21066t + −2173534e−2173534t e−21066t e−2173534t e−2152468t lne−2152468t t Vout Vout

= 0 = = = = = =

21066 2173534 9.6920 × 10−3 ln9.6920 × 10−3 2.1540µs 290.3625(e−21066(2.1540µs) − e−2173534(2.1540µs) )kV 274.78095kV

Vout × 100% Vin 274.08795kV × 100% = 300kV = 91.59365%

η =

8

3.2

Matlab® simulations

Figure 4: Circuit design with Matlab® Simulink

Figure 5: Wavefrom generated by Matlab® Simulink

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3.3

the Matlab program to compute for lightning impulse wave form

This program Calculates Front time and tail time of Lighting Impulse Generator circuit EET 4106 High Voltage Engineering r1=input(’Enter wave front resistor, R1 on IG side = ’); r2=input(’Enter wave tail resistance, R2 = ’); c1=input(’Enter capacitance of the impulse generator, C1 = ’); c2=input(’Enter capacitance of the test object, C2 = ’); Vin=input(’Enter the charging voltage of impulse generator, Vin = ’); a = (1/(r1*c2))+(1/(r2*c2))+(1/(r1*c1)); b = 1/(r1*r2*c1*c2); c = r1*c2; alpha = a/2; beta = sqrt((a/2)^2 - b); y1 = alpha - beta; y2 = alpha + beta; tf = (1/(y2-y1))*log(y2/y1); K = 0.7/((y1)*tf); tt = K*tf; A=(Vin/(c*(y2-y1))); step=0.1e-6; t(1) = 0; for i = 1 : 1000 e(i) = A*(exp(-y1*t(i))-exp(-y2*t(i))); t(i+1) = t(i) + step; end t = t(1:length(t)-1); plot(t,e);grid; title(’LIGHTNING IMPULSE VOLTAGE WAVEFORM’); xlabel(’Time in seconds’); ylabel(’Magnitude’); Vp = max(e); disp(sprintf(’a = (1/(r1c2) + 1/(r2c2) + 1/(r1c1)) = \%g’,a)) disp(sprintf(’b = (1/r1r2c1c2) = \%g’,b)); disp(sprintf(’y1 = \%g’,y1)); disp(sprintf(’y2 = \%g’,y2)); disp(sprintf(’K = \%g’,K)); 10

disp(sprintf(’TIME TO FRONT, Tf = \%g sec’,tf)); disp(sprintf(’TIME TO TAIL, Tt = \%g sec’,tt)); disp(sprintf(’Impulse Peak Waveform Voltage, Vp = \%f’, Vp));hold on; Vtp = 0.5*Vp; for i = 1 : 1000 hp(i) = Vtp; end plot(t,hp,’.r’); for i = 1 : 1000 if e(i) == Vtp if t(i) > 2e-6 tt = t(i); disp(sprintf(’Tail time of the impulse wave, Tt = %g’, tt)); end end end

Figure 6: Lightning Impulse Voltage Wavefrom generated by Matlab® Simulink

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3.4

Analysis and verification of the result 1 1 1 + + R1 C1 R2 C1 R1 C2 1 1 1 = + + (1600)(7nF ) (6500)(7nF ) (1600)(300pF ) = 2.1946 × 106

α =

α 2 = 1.0973 × 106

α =

1 R1 R2 C1 C2 1 = 2.184 × 10−11 = 4.5788 × 1010

β =

√ α2 − b β = √ = 1.0973 × 1062 − 4.5788 × 1010 = 1.0762 × 106

α1 = α − β = 1.0973 × 106 − 1.0762 × 106 = 21100

α1 = α − β = 1.0973 × 106 − 1.0762 × 106 = 21100

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α2 = α + β = 1.0973 × 106 + 1.0762 × 106 = 2173500

"

#

Vo Vpeak = {e−α1 tf − e−α2 tf } 2R1 C2 β = [235.76k]{e−21100tf − e−2173500tf dVpeak (tf ) d = [235.76k]{e−21100tf − e−2173500tf = 0 dtf dtf

−e−21100tf e−2152400tf lne−2152400tf tf

= = = =

e−2173500tf 9.7078 × 103 ln9.7078 × 103 2.1533 × 10−6 −6 )

Vpeak = [235.76k]{e−21100(2.1533×10 = 223.1051kV

Vout × 100% Vin 223.1051 × 100% = 300 = 74.3684%

η =

13

−6 )

− e−2173500(2.1533×10

}